Paper: CSIR-NET December 2011 · Subject: Physical Chemistry · Chapter: Thermodynamics · Topic: Laws State Functions · Marks: 2 · Difficulty: Medium
For the liquid $\rightleftharpoons$ vapour equilibrium of a substance $\mathrm{dP} / \mathrm{dT}$ at 1 bar and 400 K is $8 \times 10^{-3} \mathrm{bar} \mathrm{K}^{-1}$. If the molar volume in the vapour form is $200 \mathrm{Lmol}^{-1}$ and the molar volume in the liquid form is negligible, the molar enthalpy of vapourisation is (1.0 bar $\mathrm{L}=100 \mathrm{J}$)
(a)$640 \mathrm{kJ} \mathrm{mol}^{-1}$
(b)$100 \mathrm{kJ} \mathrm{mol}^{-1}$
(c)$80 \mathrm{kJ} \mathrm{mol}^{-1}$
(d)64 kJ mol
Answer
Answer: D ✓ checked by 4AB · confidence high
Explanation
Clapeyron: ΔH_vap = T·ΔV·(dP/dT) = 400 K × 200 L mol⁻¹ × 8×10⁻³ bar K⁻¹ = 640 bar L mol⁻¹ = 64,000 J = 64 kJ mol⁻¹.